Arrange the following complexes in increasing order of their magnetic moment
(A) \( [NiCl_4]^{2-} \)
(B) \( [Fe(H_2O)_6]^{2+} \)
(C) \( [Cr(H_2O)_6]^{3+} \)
(D) \( [Co(C_2O_4)_3]^{3-} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (D), (A), (C), (B)
The spin-only magnetic moment depends on the number of unpaired electrons, n:
\( \displaystyle \mu = \sqrt{n(n+2)}\ \mathrm{BM} \)
For each complex, the oxidation state of the metal comes from the charges, the d electron count from Z, and the pairing from the ligand: weak field ligands (\( \ce{Cl-} \), \( \ce{H2O} \)) leave the electrons unpaired as far as possible, while a stronger field can force pairing.
(A) \( \ce{[NiCl4]^2-} \): x + 4(−1) = −2, so x = +2. Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Ni^2+} \): \( \mathrm{3d^{8}} \). The complex is tetrahedral (\( \mathrm{sp^{3}} \)), with 2 unpaired electrons: \( \mu = \sqrt{8} = 2.83 \) BM.
(B) \( \ce{[Fe(H2O)6]^2+} \): Fe is +2. Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \); losing the two 4s electrons gives \( \mathrm{3d^{6}} \). Water is a weak field ligand, so the complex is high spin with 4 unpaired electrons: \( \mu = \sqrt{24} = 4.90 \) BM.
(C) \( \ce{[Cr(H2O)6]^3+} \): Cr is +3. Cr (Z = 24) is \( \mathrm{[Ar]\,3d^{5}\,4s^{1}} \); losing the 4s electron and two 3d electrons gives \( \mathrm{3d^{3}} \): 3 unpaired electrons whatever the ligand, \( \mu = \sqrt{15} = 3.87 \) BM.
(D) \( \ce{[Co(C2O4)3]^3-} \): x + 3(−2) = −3, so x = +3. Co (Z = 27) is \( \mathrm{[Ar]\,3d^{7}\,4s^{2}} \); losing the two 4s electrons and one 3d electron gives \( \mathrm{3d^{6}} \). With oxalate the splitting is larger than the pairing energy for \( \ce{Co^3+} \), so all six electrons pair in the lower orbitals (\( t_{2g}^{6} \)): 0 unpaired electrons, \( \mu = 0 \) (diamagnetic).
In increasing order of magnetic moment:
\( \displaystyle \ce{[Co(C2O4)3]^3-}\,(0) \lt \ce{[NiCl4]^2-}\,(2.83) \lt \ce{[Cr(H2O)6]^3+}\,(3.87) \lt \ce{[Fe(H2O)6]^2+}\,(4.90) \)
Shortcut: Compare unpaired electrons only (0, 2, 3, 4); the formula need not be worked out. The diamagnetic (D) comes first and the high spin \( \mathrm{d^{6}} \) (B) last, which only Option (2) has.